 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008-printed-variant/p/p4/a/solution

A = QR

, where

Q^{T} Q = I

and

R

is upper triangular. If

A

is orthogonal, so is

R = Q^{T} A

. Its inverse is upper triangular, but

R^{- 1} = R^{T}

is lower triangular, forcing

R

R = diag (ε_{1}, \dots, ε_{n}), ε_{i} \in {1, - 1} .

With the customary positive diagonal convention,

R = I

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